Problem involving fundamental principle

[Samshorn]

But minus the classical Lagrangian divided by the particle's mass also happens to be the equation for the elapsed proper time of the particle following the trajectory h(t)...

No it isn't. You'd see this if you actually tried to go from the metric line element to that classical Lagrangian. Please note that changing the sign and dividing by the particle's mass are not the issue. Those are trivial. The issue is that the proper time is the square root of the quadratic line element, whereas the Lagrangian is essentially proportional to the quadratic line element itself - not to the square root. So they are not the same things, nor are they even proportional to each other. So it's incorrect to claim that the classical Lagrangian represents d tau.

That can be shown directly from the metric.

No it can't, because it's false. What CAN be shown is that extremizing the classical Lagrangian yields (approximately) the same paths as maximizing the elapsed proper time, but this does not imply that the classical Lagrangian is proportional to the elapsed proper time (which it isn't). And of course, showing that these two things yield the same trajectories is precisely what is needed here, and it's precisely what we do when we derive the geodesic equations from the metric. By leaving out this derivation, and simply taking for granted that the paths yielded by the classical Lagrangian are the ones that maximize proper time, you are simply assuming the very thing you've been asked to demonstrate.

Please go back and re-read all my posts. You are correct that the expression gh - 1/2 v^{2} is not the classical Lagrangian, strictly speaking. It's minus the classical Lagrangian divided by the particle's mass. I said that explicitly in at least one previous post, and it seems like the OP understands the distinction.

Again, that's not the issue. The difference between the metric expression for d tau and the classical Lagrangian for a free-falling particle is greater than you realize. You would see this if you actually try to go from one to the other. Furthermore, by assuming the classical Lagrangian of a free particle in a gravitational field with acceleration g, you have already assumed the solution. I mentioned this in my first post. The definition of "g", the acceleration of gravity, is d^2r/dt^2, so this already gives the equation of the trajectory.

You pointed out, correctly, that g can't be considered constant for the problem as specified, which I hadn't spotted because I hadn't run the actual numbers. If g can't be considered constant, then the integral of minus the Lagrangian divided by the mass gets more complicated, as I noted in my last post.

Sure, but we still have (by definition) d^2r/dt^2 = g, recognizing that g = -m/r. This is nothing but the geodesic equation (in the classical approximation). Solving this gives the cycloid trajectory. My point is that you aren't showing how this emerges from maximizing the proper time in a spherically symmetrical metric field, you are simply assuming it, and then disguising the assumption by submerging it in some irrelevant manipulations of the classical Lagrangian.

 

[PeterDonis]

The issue is that the proper time is the square root of the quadratic line element

Agreed.

whereas the Lagrangian is essentially proportional to the quadratic line element itself - not to the square root.

Can you clarify the relationship? The OP may not understand what that means.

By leaving out this derivation, and simply taking for granted that the paths yielded by the classical Lagrangian are the ones that maximize proper time, you are simply assuming the very thing you've been asked to demonstrate.

Actually, what I did has no necessary connection to the classical Lagrangian. I derived the expression I wrote down for dtau by taking the square root of the line element, as you say. I probably should have made it clearer in my original post that that was what I was doing.

Furthermore, by assuming the classical Lagrangian of a free particle in a gravitational field with acceleration g, you have already assumed the solution.

I didn't assume that at all. As I said, I derived an expression for the proper time by taking the square root of the line element. I did note that the formula I came up with was equal to minus the classical Lagrangian divided by the mass, but I'll agree that the connection between the two is not as simple as I made it seem. I did not mean to imply that I was taking the classical Lagrangian as a starting point; I wasn't. I made no use of it at all in my actual derivation.

I mentioned this in my first post. The definition of "g", the acceleration of gravity, is d^2r/dt^2, so this already gives the equation of the trajectory.

That's not the definition of g, at least not if you do the derivation by taking the square root of the line element; "g" is then GM / R^{2}, where R is the radius of the Earth.

My point is that you aren't showing how this emerges from maximizing the proper time in a spherically symmetrical metric field, you are simply assuming it

No, I'm not. I've explained what I did above, but just to recap quickly:

(1) I took the square root of the line element for a purely radial trajectory;

(2) I expanded the square root and retained only leading order terms;

(3) I subtracted out the constant potential at the surface of the Earth;

(4) I wrote r = R + h, where R is the radius of the Earth, and made the approximation h << R (which, as you noted, is not actually valid for the problem as stated, but would be valid for a short enough trajectory);

(5) I wrote "g" for GM / R^{2};

(6) I left out the constant factor 1 / c^{2} in the formulas. This changes the units to "energy per unit mass", or "action per unit mass" when integrated with respect to time, but doesn't change anything else;

(7) I wrote down the integral of the resulting expression with respect to time, and applied the Euler-Lagrange equation to it, which finds the function h(t) that maximizes the integral. This will be the function h(t) that maximizes the proper time; and to find the actual proper time, the units will need to be corrected and the (constant) time elapsed on the ground clock (which I effectively subtracted off by subtracting the constant potential at the Earth's surface) will need to be added back.

Nowhere in the above did I use the classical Lagrangian; I merely commented on the similarities of the formula I obtained to the classical Lagrangian. I just did a derivation from the line element. If you want to say that's a roundabout way of deriving the geodesic equation, that's fine. But it's certainly not the same as *assuming* the geodesic equation or the classical Lagrangian; the only thing I *assumed* was the metric.

 

[Samshorn]

For ease of reference, here's what you wrote in post #4, where you introduced your derivation:

The full equation for proper time in a gravitational field is rather complicated, but the following approximate formula, which is derived from the exact equation (which comes from General Relativity), should work fine for this problem:

τ = ∫dτ = ∫ [gh − 1/2(dh/dt)^2]dt

where h is your height above the ground (as a function of the time t elapsed on the ground clock), g is the acceleration due to gravity, and T is the time elapsed on the ground clock when you are supposed to return (1 hour)--you're assumed to start at time 0.

So you clearly defined, or at least stated, that g is the acceleration of gravity. Also, there's no indication of how you go from the expression for proper time to this expression (which I think is what the OP was asking for), you just asserted it. Obviously once you get to that expression, which is the classical Lagrangian for a free-falling particle, and have identified g as the acceleration of gravity, i.e., d^2h/dt^2 = -g, you're done. That's the geodesic equation for the trajectory.

I derived the expression I wrote down for dtau by taking the square root of the line element, as you say. I probably should have made it clearer in my original post that that was what I was doing... If you do the derivation by taking the square root of the line element; "g" is then GM / R^{2}, where R is the radius of the Earth.

Well, you stated in your original post that your "g" represented the acceleration of gravity (see above), so it's understandable that people would think that's what you meant.

By the way, isn't there something odd about your use of "g" (=m/r^2) here? You supposedly derived the expression from the square root of the metric, but that gives the factor (1 - 2m/r)^(1/2), and hence to the first approximation 1 - m/r, and the variable part is just -m/r. At no point does the quantity m/r^2 appear in this derivation. In other words, the metric gives the gravitational potential directly as -m/r, and yet you wrote the gravitational potential not as -m/r, but rather as g*r, where you defined g = -m/r^2. It's hard to see why anyone would write it like that if they were deriving it from the metric. Do you see what I mean?

And when you go on to call "g" the acceleration of gravity, well... Also, the use of h instead of r is strange, as is the fact that you omitted any discussion of deriving it from the square root of the metric to maximize proper time until just now, even though that's precisely what the OP requested...

I've explained what I did above, but just to recap quickly:

I don't disagree with what you outlined there, other than the use of the past tense. What you're describing now is not what you presented previously. (See the quote above.) Essentially what you're describing now is what I outlined for deriving the geodesic equation from the metric line element, although you stop short of actually deriving the cycloid solution.

If you want to say that's a roundabout way of deriving the geodesic equation, that's fine.

Yes, what you're outlining now is a roundabout way of deriving the geodesic equation from the condition of maximizing the proper time. What you presented previously wasn't.

 

[PeterDonis]

So you clearly defined, or at least stated, that g is the acceleration of gravity.

Which is equal to GM/R^2 at the surface of the Earth. I already agreed that I should have been clearer about the initial presentation. See further comments below.

Obviously once you get to that expression, which is the classical Lagrangian for a free-falling particle

Only it isn't; it's minus the classical Lagrangian divided by the mass of the particle. And you yourself have already pointed out that the relationship of this expression with the classical Lagrangian is not as simple as I made it seem.

By the way, isn't there something odd about your use of "g" (=m/r^2) here? You supposedly derived the expression from the square root of the metric, but that gives the factor (1 - 2m/r)^(1/2), and hence to the first approximation 1 - m/r, and the variable part is just -m/r.

Yes, but r is not height above the Earth's surface; it's distance from the Earth's center. The equation I wrote down has h, height above the Earth's surface. As you noticed:

Also, the use of h instead of r is strange

Not if you're trying to answer the question Feynman actually asked: how to maximize the proper time on the rocket clock, given that an hour elapses *on the ground clock*. The integral I wrote down has "t" as the time of the ground clock, *not* Schwarzschild coordinate time. That means you need to adjust the "zero" of the potential to the radius of the Earth, which is the radius of the ground clock; and it also means you want to change variables from r to h, so that your distance variable is distance from the ground. That's what steps 3 and 4 in what I posted last time are for. When you do all that, you end up with a term GM/R^2 times h; so the "g" pops out, and you might as well recognize that. That's what step 5 in what I posted last time is for.

as is the fact that you omitted any discussion of deriving it from the square root of the metric to maximize proper time until just now, even though that's precisely what the OP requested...

I thought the OP was asking about the maximization part, not the part about how you derive the initial expression that then gives you the integral to maximize. I would be glad to post more details about how the equation I wrote down gets derived from the metric if the OP requests it.

Yes, what you're outlining now is a roundabout way of deriving the geodesic equation from the condition of maximizing the proper time. What you presented previously wasn't.

What I outlined in my last post *is* what I presented previously, just with a clearer explanation of the steps. See above.

 

[Samshorn]

What I outlined in my last post *is* what I presented previously, just with a clearer explanation of the steps. See above.

No it isn't. Remember you wrote

τ = ∫dτ = ∫ [gh − 1/2(dh/dt)^2]dt

but now you agree (I think) that that isn't right. That expression isn't the elapsed proper time along the trajectory, it's an approximation for the DIFFERENCE between the proper time along the trajectory and the proper time on the ground. So the expression you were insisting represented dtau really doesn't at all. Also, as noted before, you defined g as the acceleration of gravity, thereby begging the question.

Which is equal to GM/R^2 at the surface of the Earth.

Remember the thread of the conversation. ME: "You said g was the acceleration of gravity". YOU: 'I said no such thing. g is -m/r^2." ME: "Well, here's a quote where you said g was the acceleration of gravity". YOU: "Yes, which equals -m/r^2". And so it goes.

The point is that you said g was the acceleration of gravity, which means d^2h/dt^2 = g for the object in question, which is already the geodesic equation.

Not if you're trying to answer the question Feynman actually asked: how to maximize the proper time on the rocket clock, given that an hour elapses *on the ground clock*.

To the order of approximation that you are working, the weak-slow limit, I think that distinction is insignificant. In other words, the height of the apogee that you compute is going to be essentially the same, whether you say the trip was 1 hour on the Earth clock or 1 hour on the ballistic clock. If you really were trying to work it out exactly, you would have to do it the way I described, rather than making all those weak-slow approximations.

 

[Alcubierre]

Actually, if the problem covers enough range of height that "g" changes (which this one does, I apologize for not pointing that out earlier), then g itself has to be treated as a function.

He does say he was wrong in some sense. The point of this thread was to maximize the programs of height and velocity and calculate what the proper time would be and I still haven't been able to do that because of the discord between you two. Peter provided me with a good idea on how to approach this (and even taught me, indirectly, how to solve Euler-Lagrange equations) and went through the steps enough so that I could do some thinking and work myself and still be able to solve it. Sam, you say your approach is better than Peter's "weal-slow limit," and it may be the case because you use a more explicit approach, but Peter did have in mind that I only have a single-variable calculus background and he was trying to make it easy for me to understand. I'm trying my hardest here to learn what I must to solve these problems, but giving a qualitative response as to how to solve isn't helping me much because I don't know what to do with your response below but look them up on Wikipedia or other sources and still be lost,

From the equivalence principle find the field equations as the simplest covariant equations involving derivatives of no greater than second order. From the field equations derive the metric for a spherically symmetrical field (to represent the Earth's gravity). By definition, the timelike geodesics of this manifold maximize (or extremize) proper time. Apply the calculus of variations to the metric to determine the equations of geodesic paths in this manifold. Then consider a purely radial geodesic path, and integrate the geodesic equations to show that the radial position versus proper time is a cycloid (coincidentally, just as it is in Newtonian theory). Then plug in the conditions of the stated problem (one hour round trip from the Earth's surface) to determine the parameters of the trajectory.

If you would show somewhat what you did so I could get started, that'd be great for me and for Peter to see how you did it yourself, if he hasn't already.

 

[PeterDonis]

Remember you wrote

τ = ∫dτ = ∫ [gh − 1/2(dh/dt)^2]dt

but now you agree (I think) that that isn't right. That expression isn't the elapsed proper time along the trajectory, it's an approximation for the DIFFERENCE between the proper time along the trajectory and the proper time on the ground.

Yes, that's true, and also the units are wrong; there should be a 1/c^2 in front. So the actual elapsed proper time would be:

\tau = T + \int d\tau = T + \int_{0}^{T} \frac{1}{c^{2}} \left[ gh - \frac{1}{2} \left( \frac{dh}{dt} \right)^{2} \right] dt

where T = 1 hour is the time elapsed on the ground clock, as before. For solving the maximization problem that was posed in the OP, these differences don't matter; you still end up with the expression in brackets as what you need to apply the Euler-Lagrange equation to to obtain the function h(t) that maximizes the proper time. That's what I was focused on in my earlier posts.

Also, as noted before, you defined g as the acceleration of gravity

I agree my wording wasn't clear in my original post. I did not mean that I just inserted g = acceleration of gravity in the equation I wrote down. I meant that in the course of deriving the equation I wrote down from the metric, the expression GM/R^2 appeared, which I then wrote "g" in place of, since it obviously equaled the acceleration of gravity. But that's not the same as just inserting g "from nowhere" in the equation. I didn't do that. I apologize if it wasn't clear from my original post that I didn't do that, but the fact remains that I didn't. I described how I derived my original equation from the metric, without making any assumptions about "g".

To the order of approximation that you are working, the weak-slow limit, I think that distinction is insignificant. In other words, the height of the apogee that you compute is going to be essentially the same, whether you say the trip was 1 hour on the Earth clock or 1 hour on the ballistic clock.

This is correct, but it misses my point. The "t" in the Schwarzschild metric is Schwarzschild *coordinate* time; but in the statement of the problem we are given time on the ground clock, not coordinate time. So I corrected the potential to be zero at h = 0, i.e., at the surface of the Earth. That let's the limits of integration be specified directly in terms of the given: 1 hour elapsed on the ground clock.

 

[PeterDonis]

Alcubierre, actually a fair bit of what Samshorn is describing is just to get to an equation like the one I originally wrote down (which only applies if the time is short enough that g can be considered constant), or a more complicated version when g is not constant, like what I wrote down in post #20. Basically there are three stages to the overall process, and I have only been talking about the third.

The three stages are:

(1) Determine the general physical law that governs gravity. This is what Samshorn is talking about here:

From the equivalence principle find the field equations as the simplest covariant equations involving derivatives of no greater than second order.

The "field equations" he is talking about are the Einstein Field Equations, which are the central equations of General Relativity. I was taking those as given for the purposes of this problem, but you can get a quick overview about them here:

http://en.wikipedia.org/wiki/Einstein_field_equations

(2) From the general physical law that governs gravity, find the particular solution that describes gravity around a spherically symmetric body (which we can approximate the Earth to be for this problem). This is what Samshorn is talking about here:

From the field equations derive the metric for a spherically symmetrical field (to represent the Earth's gravity).

The "metric" Samshorn refers to is called the Schwarzschild metric, and I was also taking it as given for purposes of this problem, but you can read more about it here:

http://en.wikipedia.org/wiki/Schwarzschild_metric

(3) Once you have the solution describing gravity around a spherically symmetric body, use it to determine what trajectory--what function h(t)--maximizes the time on the rocket's clock. This is what Samshorn is talking about here:

By definition, the timelike geodesics of this manifold maximize (or extremize) proper time. Apply the calculus of variations to the metric to determine the equations of geodesic paths in this manifold. Then consider a purely radial geodesic path, and integrate the geodesic equations to show that the radial position versus proper time is a cycloid (coincidentally, just as it is in Newtonian theory). Then plug in the conditions of the stated problem (one hour round trip from the Earth's surface) to determine the parameters of the trajectory.

What I walked you through was an approximate version of the above, valid only if g can be considered constant. I'll let Samshorn give more specifics/references about the equations he would use for the case when g is not constant. He pointed out, correctly, that for this problem the Newtonian equations, which are simpler, are sufficient.

 

[Alcubierre]

Hello again, guys. I had some long personal businesses to take care, now I'm back and working on this again. So, Peter to be clear, I should maximize the integrand on post #37 or use a Newtonian equation? Also, I was having difficulties integrating that. I will post more about it tomorrow.

 

[PeterDonis]

So, Peter to be clear, I should maximize the integrand on post #37 or use a Newtonian equation?

Remember that the integral I wrote down in #37 is valid only for the case where g can be considered constant. As Samshorn pointed out a number of posts ago, for the specific case in question (elapsed time of one hour for the clock on the ground), g *can't* be considered constant, so the integral gets more complicated.

For the case where g can be considered constant, yes, you would maximize the integral in #37, which actually amounts to maximizing the expression in brackets--more precisely, you would find the function h(t) that maximizes the expression in brackets.

I'm not sure what you mean by "use a Newtonian equation".

 

[Alcubierre]

I still have not been able to figure this out. I stopped working on it for a while because I've been extremely busy. So to recap, when I apply calculus of variation (Euler-Lagrange) to the integrand on post #37, I get a(t) which I can integrate once to obtain v(t) and once more to obtain h(t). Then what do I do with those equations in respect to time, simply plug in a value for t or what? That's the part that is throwing me off a bit.

 

[PeterDonis]

when I apply calculus of variation (Euler-Lagrange) to the integrand on post #37, I get a(t) which I can integrate once to obtain v(t) and once more to obtain h(t). Then what do I do with those equations in respect to time, simply plug in a value for t or what?

If you want to answer the question posed in the OP (in the Feynman quote), the function h(t) *is* the answer.

If you want to find the actual numerical value of the height at a given time t, yes, plug that value of t into the function h(t). Note that, as I said in an earlier post, h(t) will have a term v_0 t, as well as the - 1/2 g t^2 term, corresponding to an initial upward velocity v_0. To find the actual numerical value of v_0, you use the fact that h(t) = 0 when t = 1 hour; that let's you solve for v_0 in terms of g.

 

[Alcubierre]

So what is the difference between using 1/(c^2)[gh - .5gh'^2] (from post #37) and GM/(R^2) - .5gh'^2?

 

[PeterDonis]

So what is the difference between using 1/(c^2)[gh - .5gh'^2] (from post #37) and GM/(R^2) - .5gh'^2?

I'm not sure what you actually meant to ask about with the second formula. If you meant this (note that I have inserted an h in the first term and a factor of 1 / c^2 in front to make the units the same as those in the first formula)...

\frac{1}{c^2} \left[ \frac{GM}{R^2} h - \frac{1}{2} \left( \frac{dh}{dt} \right)^2 \right]

...then this is just a rewrite of the first formula with GM/R^2 substituted for g. But if you meant the alternate formula I gave in post #20, you've left out a term and got an extra factor of R in the first term; it should be (correcting the units again):

\frac{1}{c^2} \left[ \frac{GM}{R} - \frac{GM}{R + h} - \frac{1}{2} \left( \frac{dh}{dt} \right)^2 \right]

If the latter is what you meant, the difference between the two formulas is that the first is only valid for short trajectories, while the second is valid generally. As Samshorn pointed out in a fairly early post in this thread, a one-hour time of flight, which is the actual number posed in the OP, is long enough that the first formula is not valid. However, as I pointed out in a somewhat later post, integrating the second formula is harder because of the h in the denominator of the second term.